Editing Uniform convergence (section)

===To continuity===
{{Main|Uniform limit theorem}}
[[Image:Drini nonuniformconvergence SVG.svg|thumb|350px|right|Counterexample to a strengthening of the uniform convergence theorem, in which pointwise convergence, rather than uniform convergence, is assumed. The continuous green functions <math>\sin^n(x)</math> converge to the non-continuous red function. This can happen only if convergence is not uniform.]]

If <math>E</math> and <math>M</math> are [[topological space|topological spaces]], then it  makes sense to talk about the [[continuous function (topology)|continuity]] of the functions <math>f_n,f:E\to M</math>.  If we further assume that <math>M</math> is a [[metric space]], then (uniform) convergence of the <math>f_n</math> to <math>f</math> is also well defined.  The following result states that continuity is preserved by uniform convergence:

{{math theorem | name = Uniform limit theorem | math_statement = Suppose <math>E</math> is a  topological space, <math>M</math> is a metric space, and <math>(f_n)</math> is a sequence of continuous functions <math>f_n:E\to M</math>.  If <math>f_n \rightrightarrows f</math> on <math>E</math>,  then <math>f</math> is also continuous.}}

This theorem is proved by the "{{math|&epsilon;/3}} trick", and is the archetypal example of this trick: to prove a given inequality ({{math|&epsilon;}}), one uses the definitions of continuity and uniform convergence to produce 3 inequalities ({{math|&epsilon;/3}}), and then combines them via the [[triangle inequality]] to produce the desired inequality.

{{Proof|Let <math> x_0\in E </math> be an arbitrary point. We will prove that <math> f</math> is continuous at <math>x_0</math>. Let <math>\varepsilon >0 </math>. By uniform convergence, there exists a natural number <math>N</math> such that

<math>\forall x \in E\quad d(f_N(x),f(x))\leq \tfrac{\varepsilon}{3}</math>

(uniform convergence shows that the above statement is true for all <math>n\geq N</math>, but we will only use it for one function of the sequence, namely <math>f_N</math>).

It follows from the continuity of <math> f_N</math> at <math> x_0\in E</math> that there exists an [[open set]] <math> U</math> containing <math> x_0</math> such that

<math>\forall x\in U\quad d(f_N(x),f_N(x_0))\leq\tfrac{\varepsilon}{3}</math>.

Hence, using the [[triangle inequality]],

<math>\forall x\in U\quad d(f(x), f(x_0))\leq d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))\leq\varepsilon</math>,

which gives us the continuity of <math> f</math> at <math> x_0</math>.<math>\quad\square</math>}}

This theorem is an important one in the history of real and Fourier analysis, since many 18th century mathematicians had the intuitive understanding that a sequence of continuous functions always converges to a continuous function.  The image above shows a counterexample, and many discontinuous functions could, in fact, be written as a [[Fourier series]] of continuous functions.  The erroneous claim that the pointwise limit of a sequence of continuous functions is continuous (originally stated in terms of convergent series of continuous functions) is infamously known as "Cauchy's wrong theorem".  The uniform limit theorem shows that a stronger form of convergence, uniform convergence, is needed to ensure the preservation of continuity in the limit function.

More precisely, this theorem states that the uniform limit of ''[[uniformly continuous]]'' functions is uniformly continuous; for a [[locally compact]] space, continuity is equivalent to local uniform continuity, and thus the uniform limit of continuous functions is continuous.